Lemma 10.136.18. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a syntomic map. Then there exists elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i} \cong S_ i/IS_ i$ for some relative global complete intersection $S_ i$ over $R$.

**Proof.**
By Lemma 10.136.15 we find a collection of elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i}$ is a relative global complete intersection over $R/I$. Hence we may assume that $\overline{S}$ is a relative global complete intersection. Write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ c)$ as in Definition 10.136.5. Choose $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$. Set $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Note that $S/IS \cong \overline{S}$. By Lemma 10.136.10 we can find $g \in S$ mapping to $1$ in $\overline{S}$ such that $S_ g$ is a relative global complete intersection over $R$. Since $\overline{S} \cong S_ g/IS_ g$ this finishes the proof.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: